Question

If the distance from the points (6,-2) and (3,4) to the lines 4x+3y=12,4x-3y=12 are d1 and d2 respectively then d1:d2

Answer 1

SOLUTION

GIVEN

The distance from the points (6,-2) and (3,4) to the lines 4x+3y=12 , 4x-3y=12 are

$\sf{d_1 \: \: and \: \: d_2}$ respectively

TO DETERMINE

$\sf{d_1 : d_2}$

CONCEPT TO BE IMPLEMENTED

Distance of the point $\sf{(x_1, y_1)}$

from the line ax + by + c = 0 is

$\displaystyle \sf{d = \bigg| \: \frac{ax_1 + by_1 + c}{ \sqrt{ {a}^{2} + {b}^{2} } } \: \bigg| \: \:}unit$

EVALUATION

Here the given given lines are 4x+3y =12 , 4x-3y=12

Now The distance from the point (6,-2) to the line 4x+3y=12 is

$\displaystyle \sf{d_1 = \bigg| \: \frac{(4 \times 6) + (3 \times - 2) - 12}{ \sqrt{ {4}^{2} + {3}^{2} } } \: \bigg| \: \:}unit$

$\displaystyle \sf{ \implies \: d_1 = \bigg| \: \frac{24 - 6 - 12}{ \sqrt{ 16 + 9 } } \: \bigg| \: \:}unit$

$\implies \displaystyle \sf{d_1 = \frac{6}{5} \: \:}unit$

Now The distance from the point (3,4) to the line 4x-3y=12 is

$\displaystyle \sf{d_2 = \bigg| \: \frac{(4 \times 3) - (3 \times 4) - 12}{ \sqrt{ {4}^{2} + {( - 3)}^{2} } } \: \bigg| \: \:}unit$

$\displaystyle \sf{ \implies \: d_2 = \bigg| \: \frac{12 - 12- 12}{ \sqrt{ 16 + 9 } } \: \bigg| \: \:}unit$

$\displaystyle \sf{ \implies \: d_1 = \frac{12}{5} \: \:}unit$

$\therefore \: \: \sf{d_1 : d_2}$

$= \displaystyle \sf{ \frac{6}{5} : \frac{12}{5} \: \:}$

$= 1 : 2$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. find the equation of the line passing through the point of intersection of the lines 5x - 8y + 23 = 0 and 7x + 6y - 71 =0

https://brainly.in/question/22027728

2. The angle between the lines x + 2y = 11 and 2x – y = 9 is

https://brainly.in/question/29875521